CARDS
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 1448 | Accepted: 773 |
Description
The shuffle machine accepts the set of cards arranged in an arbitrary order and performs the following operation of double shuffle : for all positions i, 1 <= i <= N, if the card at the position i is j and the card at the position j is k, then after the completion of the operation of double shuffle, position i will hold the card k.
Alice and Bob play a game. Alice first writes down all the numbers from 1 to N in some random order: a1, a2, ..., aN. Then she arranges the cards so that the position ai holds the card numbered ai+1, for every 1 <= i <= N-1, while the position aN holds the card numbered a1.
This way, cards are put in some order x1, x2, ..., xN, where xi is the card at the ith position.
Now she sequentially performs S double shuffles using the shuffle machine described above. After that, the cards are arranged in some final order p1, p2, ..., pN which Alice reveals to Bob, together with the number S. Bob's task is to guess the order x1, x2, ..., xN in which Alice originally put the cards just before giving them to the shuffle machine.
Input
The following N lines describe the final order of cards after all the double shuffles have been performed such that for each i, 1 <= i <= N, the (i+1)st line of the input file contains pi (the card at the position i after all double shuffles).
Output
For each i, 1 <= i <= N, the ith line of the output file should contain xi (the card at the position i before the double shuffles).
Sample Input
7 4
6
3
1
2
4
7
5
Sample Output
4
7
5
6
1
2
3
Source
Mean:
剀剀和凡凡有N张牌（依次标号为1，2，……，N）和一台洗牌机。假设N是奇数。洗牌机的功能是进行如下的操作：对所有位置I（1≤I≤N），如果位置I上的牌是J，而且位置J上的牌是K，那么通过洗牌机后位置I上的牌将是K。
剀剀首先写下一个1~N的排列ai，在位置ai处放上数值ai+1的牌，得到的顺序x1, x2, ..., xN作为初始顺序。他把这种顺序排列的牌放入洗牌机洗牌S次，得到牌的顺序为p1, p2, ..., pN。
现在，剀剀把牌的最后顺序和洗牌次数告诉凡凡，要凡凡猜出牌的最初顺序x1, x2, ..., xN。
analyse:
刚开始搞置换群，看得云里雾里的，还好看到了潘震皓的《置换群快速幂运算 + 研究与探讨》，讲的很清楚，而且很符合ACM的出题习惯。
很显然，这是一题典型的置换群问题，一副扑克就是一个置换，而对于每次的操作，我们可以看作置换的平方运算，题目说n为奇数，这就保证了在进行置换平方运算的过程中不会出现分裂，那么我们就可以使用置换群的快速幂来做了。
进行2*x次运算就可，当然其中有一个剪枝，将O(n^2+logs)的时间复杂度变为了O(n+logs)了，十分经典。
Time complexity：O(n+logs)
Source code：
// Memory Time
// 1347K 0MS
// by : Snarl_jsb
// 2014-09-11-20.34
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define N 1010
#define LL long long
using namespace std;
int a[N],b[N],c[N],n,m;
int work()
{
int j;
int cnt=0;
while(1)
{
for(int i=1;i<=n;i++)
b[i]=c[c[i]];
cnt++;
for(j=1;j<=n;j++)
if(b[j]!=a[j])
break;
if(j>n)break;
for(int i=1;i<=n;i++)
c[i]=b[i];
}
return cnt;
}
int main()
{
// freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);
// freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
c[i]=a[i];
b[i]=a[i];
}
int cnt=work();
m%=cnt;
m=cnt-m;
while(m--)
{
for(int i=1;i<=n;i++)
b[i]=a[a[i]];
for(int i=1;i<=n;i++)
a[i]=b[i];
}
for(int i=1;i<=n;i++)
printf("%d\n",b[i]);
}
return 0;
}